I Can a Basis Vector be Lightlike?

[cianfa72]

[Moderator's note: Spin off from another thread due to topic change.]

I was thinking about the following: can we take as a basis vector a null (i.e. lightlike) vector to write down the metric ?

Call $v$ such a vector and add to it 3 linear independent vectors. We get a basis for the tangent space (the first vector in the basis is the $v$ vector itself).

Then in such a basis the metric coefficient $g_{00}$ should be zero since the vector $v$ has coordinate $(1,0,0,0)$ and by definition it has zero length. Hence we are not allowed to use it, I believe.

 

[martinbn]

Sorry I was thinking about the following: can we take as a basis vector a null (i.e. lightlike) vector to write down the metric ?

Call $v$ such a vector and add to it 3 linear independent vectors. We get a basis for the tangent space (the first vector in the basis is the $v$ vector itself).

Yes, you can do that.

Then in such a basis the metric coefficient $g_{00}$ should be zero since the vector $v$ has coordinate $(1,0,0,0)$ and it has zero length. Hence we are not allowed to use it, I believe.

You are not allowed to use what?

 

[cianfa72]

You are not allowed to use what?

If we use the vector $v$ as element in the basis we get a metric in which $g_{00}$ is actually zero. Does a such metric make sense ?

 

[martinbn]

If we use the vector $v$ as element in the basis we get a metric in which $g_{00}$ is actually zero. Does a such metric make sense ?

Of course. That is just a component in a given basis, it can be zero or anything else.

 

[cianfa72]

Of course. That is just a component in a given basis, it can be zero or anything else.

BTW it does mean the metric in that basis cannot be diagonal, otherwise components in $v$ direction do not appear in the metric at all !

 

[vanhees71]

Take the most simple (1+1)-dimensional Minkowski-space example. A basis must just two linear independent vectors. Let's choose the two light-like vectors $\boldsymbol{e}_1=(1,1)$ and $\boldsymbol{e}_2=(1,-1)$ (written as component vectors wrt. a usual pseudo-Euclidean/Lorentzian basis). Then the Minkowski product of two vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ wrt. this "light-like basis" is
$$\boldsymbol{x} \cdot \boldsymbol{y}=x^1 y^1 \boldsymbol{e}_1 \cdot \boldsymbol{e}_1 + (x^1 y^2 + x^2 y^1) \boldsymbol{e}_1 \boldsymbol{e}_2 + x^2 y^2 \boldsymbol{e}_2 \cdot \boldsymbol{e}_2 = 2 (x^1 y^2+x^2 y^1).$$

 

[cianfa72]

Take the most simple (1+1)-dimensional Minkowski-space example. A basis must just two linear independent vectors. Let's choose the two light-like vectors $\boldsymbol{e}_1=(1,1)$ and $\boldsymbol{e}_2=(1,-1)$.

Yes, indeed the (1+1) Minkowski metric in that basis is not diagonal.

 

[vanhees71]

Of course not. They can not. Think about what you can say if you have two light-like vectors that are Minkowski-orthogonal to each other! Can they be (part of) a basis?

 

[robphy]

Yes, I can see it graphically, but my problem is to show it in a free cordinate way.

You can do it in a coordinate-free way.
Given $\vec A$ and $\vec B$, then $\vec B= \vec A+ (\vec B-\vec A)$. (No coordinates were used in this calculation.)

 

[cianfa72]

Of course not. They can not. Think about what you can say if you have two light-like vectors that are Minkowski-orthogonal to each other! Can they be (part of) a basis?

Two linear independent null (light-like) vectors $\boldsymbol {u}, \boldsymbol {w}$ in Minkowski standard basis have components:

$\boldsymbol {u} = (k, k) , \boldsymbol {w} = (m , -m) \text{ } k,m \in \mathbb R$

then the further condition $\boldsymbol{u} \cdot \boldsymbol{w} = 0$ implies $km + km = 2km = 0$. Hence either $\boldsymbol {u}$ or $\boldsymbol {w}$ would result in the zero vector.

 

[vanhees71]

You can rather prove that two Minkowski-orthogonal lightlike vectors are collinear.

 

[cianfa72]

You can rather prove that two Minkowski-orthogonal lightlike vectors are collinear.

Yes, if both $\boldsymbol {u}$ and $\boldsymbol {w}$ are not the zero vector then it follows $\boldsymbol {u} = k \boldsymbol {w}$, hence they are not linearly independent (it actually includes the case at least one vector is the zero vector).

 

[ergospherical]

Also look at the textbook by Stephani et al. for usage of complex null tetrads, formed of two real null vectors $\mathbf{k}$ and $\mathbf{l}$ with $g_{kl} = \mathbf{k} \cdot \mathbf{l} = -1$ and two complex null vectors $\mathbf{m}$ and $\overline{\mathbf{m}}$ with $g_{m \overline{m}} = \mathbf{m} \cdot \overline{\mathbf{m}} = 1$. (All the other inner products vanish.)

 

[cianfa72]

Another point related to the topic. I would like to show that any 4-vector in Minkowski spacetime can be written as the sum of a timelike vector plus a spacelike one.

Take a timelike vector $\boldsymbol {u}$. It has components $(u_0,u_1,u_2,u_3)$ in Minkowski standard basis.

Consider the set of vectors $\boldsymbol {w}$ of components $(w_0,w_1,w_2,w_3)$ orthogonal to it, namely the set of vectors such that $u_0w_0 - u_1w_1 - u_2w_2 - u_3w_3=0$ in Minkowski standard basis. Such a set defines a 3-dimensional linear subspace of the vector space $V$. Furthermore each vector in such set is spacelike.

This way we get a basis of one timelike vector $\boldsymbol {u}$ and three linear independent spacelike vectors orthogonal to it. So the subspace spanned by $\boldsymbol {u}$ and the subspace $\boldsymbol {u}\perp$ are in direct sum, hence any vector in $V$ can be uniquely written as the the linear combination of vector $\boldsymbol {u}$ and a vector in $\boldsymbol {u}\perp$.

Does it make sense ? Thanks.

 

[Ibix]

Take an arbitrary vector $v^a$ and an arbitrary unit timelike vector $t^a$. Then $v_at^a$ is the component of $v^a$ parallel to $t^a$, so $s^a=v^a-v_bt^bt^a$ is the part of $v^a$ that is orthogonal to $t^a$ (note: flip the sign on the second term if your metric is -+++). It's just algebra to show that $s^a$ is a spacelike vector. Hence we have shown that $v^a=(v_bt^b) t^a+s^a$ as required.

 

[cianfa72]

Take an arbitrary vector $v^a$ and an arbitrary unit timelike vector $t^a$. Then $v_at^a$ is the component of $v^a$ parallel to $t^a$, so $s^a=v^a-v_bt^bt^a$ is the part of $v^a$ that is orthogonal to $t^a$.

$s_at^a=v_at^a - (v_bt^b)t_at^a$. Since $t^a$ is unit timelike then $s_at^a=v_at^a - v_bt^b=0$, hence $s^a$ is orthogonal to $t^a$.

 

[vanhees71]

Yes, and a vector Minkowski-orthogonal to a time-like vector is...

 

[cianfa72]

Yes, and a vector Minkowski-orthogonal to a time-like vector is...

Spacelike (if non-zero vector).

 

[Ibix]

I was assuming you'd calculate $s_as^a$ and see that it's opposite sign to $t_at^a$, but yes.

 

[cianfa72]

I was assuming you'd calculate $s_as^a$ and see that it's opposite sign to $t_at^a$, but yes.

$s_as^a = (v_a - (v_bt^b)t_a)(v^a-v_bt^bt^a) = v_av^a - (v_bt^b)v_at^a - (v_bt^b)t_av^a + (v_bt^b)^2t_at^a =$
$s_as^a = v_av^a - 2(v_bt^b)^2 + (v_bt^b)^2 = v_av^a - (v_bt^b)^2$

 

[Ibix]

Yup. From your final expression $s_as^a$ is manifestly negative if $v_av^a\leq 0$, but in the case that $v_av^a>0$ there's a tiny bit more work to do to show that $(v_av^a)\leq(v_at^a)^2$.

 

[cianfa72]

in the case that $v_av^a>0$ there's a tiny bit more work to do to show that $(v_av^a)\leq(v_at^a)^2$.

Supposing I did not any mistake before, I've no idea how to proceed .

 

[PeterDonis]

Take an arbitrary vector $v^a$ and an arbitrary unit timelike vector $t^a$.

The vectors can't be completely arbitrary since they must be linearly independent.

 

[Ibix]

The vectors can't be completely arbitrary since they must be linearly independent.

Yeah, you're right. If $v^a=kt^a$ for some constant $k$ then $s^a=0$. So you have to choose a unit timelike vector $t^a$, arbitrary except that it isn't parallel to $v^a$.

Supposing I did not any mistake before, I've no idea how to proceed .

What's $v_at^a$ in terms of $v_av^a$ and $t_at^a$, assuming $v^a$ is timelike?

 

[cianfa72]

What's $v_at^a$ in terms of $v_av^a$ and $t_at^a$, assuming $v^a$ is timelike?

Since $v_av^a > 0$ there is a Lorentz-orthonormal basis such that the components of $v^a$ and $t^a$ are $(v^0,0,0,0)$ and $(t^0,t^1,t^2,t^3)$.

In this basis $v_at^a=v^0t^0 \Rightarrow (v_at^a)^2=(v^0)^2(t^0)^2$. Since $t^a$ is unit timelike we get $(t^0)^2 > 1$ then $(v_at^a)^2 > (v^0)^2 = v_av^a$.

Note that in the above we are assuming $v^a$ and $t^a$ not collinear otherwise $s^a=0$.

 

[vanhees71]

Let $(a^{\mu})$ be a time-like future-pointing vector, i.e., $a_{\mu} a^{\mu}=(a^0)^2-\vec{a}^2>0$, $a^0>0$, and $(b^{\mu})$ a vector with $a^{\mu} b_{\mu} =a^0 b^0-\vec{a} \cdot \vec{b}=0$. Then $(b^{\mu})$ is spacelike. We have
$$b^0=\frac{1}{a^0} \vec{a} \cdot \vec{b} <\frac{1}{|\vec{a}|} \vec{a} \cdot \vec{b} \leq |\vec{b}| \; \Rightarrow \; b^0<|\vec{b}| \; \Rightarrow \; b_{\mu} b^{\mu} =(b^0)^2-\vec{b}^2<0,$$
i.e., $(b^{\mu})$ is spacelike, as claimed.

If $(a^{\mu})$ is past-pointing you can use the same argument with $(-a^{\mu})$, i.e., if $(b^{\mu})$ is Minkowski-orthogonal to an arbitrary time-like vector, it's necessarily space-like.

 

[cianfa72]

If $(a^{\mu})$ is past-pointing you can use the same argument with $(-a^{\mu})$, i.e., if $(b^{\mu})$ is Minkowski-orthogonal to an arbitrary time-like vector, it's necessarily space-like.

Why two cases (futur-poiniting and past-pointing) for the timelike vector $a^{\mu}$ ? Assuming it is timelike we always have $a_{\mu} a^{\mu}=(a^0)^2-\vec{a}^2>0$ so the above argument does not change, I believe.

 

[vanhees71]

Well, yes. You can divide by $|a^0|$ in the first step of my proof, i.e., using
$$|a^0 b^0|=|\vec{a} \cdot \vec{b}| \; \Rightarrow \; |b^0|=\frac{1}{|a^0|} |\vec{a} \cdot \vec{b}|<\frac{1}{|\vec{a}|} |\vec{a} \cdot \vec{b}| \leq |\vec{b}| \; \Rightarrow \; |b^0|^2-|\vec{b}|^2=b_{\mu} b^{\mu}<0.$$

 

[cianfa72]

Yes, see also here https://www.physicsforums.com/threads/orthogonality-of-timelike-and-null-vector.719372/post-4553594.